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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
Two elements X and Y combine to form a compound XY. Under what conditions the bond formed between them will be ionic ?A. If the difference in electronegativities of X and Y is 1.7.B. If the difference in electronegativities of X and Y is more than 1.7.C. If the difference in electronegativities of X and Y is less than 1.7.D. If both X and Y are highly electronegative. |
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Answer» Correct Answer - B For the formation of an ionic bond, the electronegativies of two atoms should differ by more than `1.7`. |
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| 852. |
The electronegativities of F, `Cl`, `Br`, and `I` are `4.0, 3.0, 2.8`, and `2.5`, respectively. The hydrogen halide with a high percentage of ionic character isA. HFB. HClC. HBrD. HI |
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Answer» Correct Answer - A Greater is the electronegativity difference between X and H, the greater is the ionic character. This ionic character of H-F is maximum |
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| 853. |
Which of the following involves `sp^2` hybridisation?A. `CO_2`B. `SO_2`C. `N_2O`D. `CO` |
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Answer» Correct Answer - B Hybrid state of S in `SO_2` is `sp^2` |
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| 854. |
A square planar complex is formed by hybridisation of which atomic orbitals?A. `s,p_x, p_y, p_z`B. `s, p_x, p_y, d_(x^2-y^2)`C. `s, p_x, p_y, d_z^2`D. `s, p_x, p_z, d_(xy)` |
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Answer» Correct Answer - B For square planar geometry, hybridisation is `dsp^2` involving s, `p_x,p_y` and `d_(x^2-y^2)` orbital. |
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| 855. |
The hybridisation of atomic orbitals of nitrogen in `NO_2^+ , NO_3^- ` and `NH_4^(+)` are :A. `sp^2, sp^3 and sp^2` respectivelyB. `sp, sp^2 and sp^3` respectivelyC. `sp^2, sp and sp^3` respectivelyD. `sp^2, sp^3 and sp` respectively |
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Answer» Correct Answer - B In `NO_2^+` total no. of valence electrons = 5 + 2 x 6-1 =16 Now 16+8 = `2(Q_1) + 0(R_1)` = 2 Thus, type of hybridization is sp. In `NO_3^(-)` total no. of valence electrons =5+ 3 x 6+1=24 Now, 24 + 8 = 3`(Q_1) + 0(R_1)` = 3 Thus type of hybridisation =`sp^2` In `NH_4^(+)` ions, total no of valence electrons = 5+1 x 4-1 =8 Now 8+2 = 4`(Q_1)+0(R_1)=4` Thus type of hybridization is `sp^3` Combining all the results, option (b) is correct. |
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| 856. |
Which of the following pairs form a stable coordinate bond ?A. NaOH. HClB. `NH_3H_2O`C. `NH_3BF_3`D. `BF_3BCl_3` |
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Answer» Correct Answer - C `NH_3` acts as a Lewis base due to the presence of lone pair which it donates to `BF_3` for the formation of a coordinate bond. `BF_3` is an electron deficient molecule. |
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| 857. |
The compound containing coordinate bond isA. one electron from an atom is transfereed to otherB. one electron each is lost from both the atomsC. a pair of electrons is contributed by one atom and shared by both the atomsD. a pair of electrons is transferred to the other atom. |
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Answer» Correct Answer - C In a coordinate or dative bond, electrons required for sharing are contributed by one atom. |
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| 858. |
Which of the following is not correct about a coordinate bond ?A. A co-ordinate bond once formed cannot be distinguished from a covalent bondB. A co-ordinate bond is also called a semi polar bondC. A co-ordinate bond is non-directional in nature.D. Due to co-ordinate bond the formal charges on donor and acceptor atoms are + and - respectively, |
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Answer» Correct Answer - C A co-ordinate bond is directional in nature just like a covalent bond. |
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| 859. |
The compound which does not contain ionic bond isA. NaOHB. HClC. `K_2S`D. LiH |
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Answer» Correct Answer - B HCl, H-atom and Cl-atom shares one electron pair to form covalent bond. |
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| 860. |
The bond order in `O_2^+` isA. 1B. 1.5C. 2.5D. 3 |
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Answer» Correct Answer - C Bond order of `O_2^+` =2.5 |
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| 861. |
The calculated bond order of superoxide ion `(O_2^(-))` isA. 0.5B. 1.5C. 3.5D. 2.5 |
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Answer» Correct Answer - A The bond order of `O_2^(-1)` is 1.5 |
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| 862. |
The bond order in `O_2^+` isA. 2B. 2.5C. 1.5D. 3 |
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Answer» Correct Answer - B M.O. configuration of `O_2^(+)` `(sigma_(1s))^2 sigma^**(1s)^2 sigma(2s)^2 sigma^**(2s)^2 sigma(2p_z)^2 pi(pi_(2p_x))^2 pi(2pi_y)^2 pi^**(2p_z)^1` Bond order =`(N_b-N_a)/2 =(10-5)/2=2.5` |
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| 863. |
Which of the following molecule has highest dipole moment?A. `H_2S`B. `CO_2`C. `C Cl_4`D. `BF_3` |
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Answer» Correct Answer - A Presence of two lone pairs on the sulphur atom, makes the geometry of `H_2S` molecule, irregular. While all the other molecules exhibit regular geometry and zero dipole moment. |
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| 864. |
The hybridisation of phosphorus in `POCl_3` is the same asA. P in `PCl_3`B. S in `SF_4`C. Cl in `ClF_3`D. B in `BCl_3` |
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Answer» Correct Answer - A P in `POCl_3` is `sp^3` hybridised i.e., H=`1/2[5+3]=4` P in `PCl_3` is also `sp^3` hybridised i.e., H=`1/2[5+3]=4` |
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| 865. |
Which of the following does not contain a coordinate bond?A. `H_3O^(+)`B. `BF_4^-`C. `HF_2^-`D. `NH_4^+` |
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Answer» Correct Answer - C `HF_2^-` has H bond `[F…. H-F]^-` |
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| 866. |
Diatomic molecule has a dipole moment of `1.2D` If its bond `1.0 Å` what fraction of an electronic charge exists on each atom ? .A. 12% of eB. 18% of eC. 25% of eD. 29% of e |
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Answer» Correct Answer - B `delta=("dipole moment")/d=(1.2D)/(1.0xx10^(-8)cm)` `=(1.2xx10^(-18) "esu.cm")/(1.0xx10^(-8)"cm")` The fraction of an electronic charge is `=(1.2xx10^(-10)"esu")/(4.8xx10^(-10) "esu/e")` |
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| 867. |
The bond order in `O_2^+` is the same as in :A. `N_2^+`B. `CN^-`C. `CO`D. `NO^+` |
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Answer» Correct Answer - A B.O. of `N_2^+=1/2[N_b-N_a]=1/2[5-0]=2.5` B.O. of `O_2^+=1/2[N_b-N_a]=1/2[6-1]=2.5` |
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| 868. |
Shape of `O_2F_2` is similar to that ofA. `C_2F_2`B. `H_2O`C. `H_2F_2`D. `C_2H_2` |
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Answer» Correct Answer - B It is an unstable orange-yellow solid. Its shape is similar to that of `H_2O_2`, except that the `O-O` bond length of `122p m` is much shorter than the `O-O` distance of `148p m` in `H_2O_2`. Both `H_2O_2` and `O_2F_2` have open book type structure. |
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| 869. |
In acetylene molecule, the carbon atoms are linked by:A. one sigma bond and two pi-bondsB. two sigma bonds and one pi-bondC. three sigma bondsD. three pi-bonds. |
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Answer» Correct Answer - A In acetylene `HC-=CH`, the two C-atoms are linked by one sigma and two pi-bonds |
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| 870. |
The `ONO` bond angle is maximum inA. `NO_3^-`B. `NO_2`C. `NO_2^-`D. `NO_2^+` |
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Answer» Correct Answer - D In `NO_2^(+)`, N atom is sp-hybridised whereas in `NO_3^(-), NO_2^(-) and NO_2`. N atom is `sp^2`-hybridized. Bond angle of sp-hybridized molecule is maximum. So `NO_2^(+)` has maximum ONO bond angle |
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| 871. |
In which pair of species, both species do have similar geometryA. `CO_2`, `SO_2`B. `NH_3`, `BH_3`C. `CO_3^(2-)`, `SO_3^(2-)`D. `SO_4^(2-),ClO_4^(-)` |
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Answer» Correct Answer - D Both `SO_4^(2-)` and `ClO_4^(-)` are tetrahedral. |
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| 872. |
The shape of `CO_2` molecule is similar toA. `H_2O`B. `BeF_2`C. `SO_2`D. none of these. |
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Answer» Correct Answer - B Both `CO_2` and `BeF_2` is linear. |
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| 873. |
Among the following species , identify the isostructural pairs . ` NF_3, NO_3^-, BF_3 , H_3O^+ , HN_3`A. `[NF_3, NO_3^(-)]` and `[BF_3 , H_3^(+)O]`B. `[NF_3, HN_3]` and `[NO_3^(-), BF_3]`C. `[NF_3, H_3^(+)O]` and `[NO_3^(-), BF_3]`D. `[NF_3, H_3^(+)O]` and `[HN_3, BF_3]` |
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Answer» Correct Answer - D `[NF_3 and H_3O^(+)]` are pyramidal while `[NO_3^(-) and BF_3]` planer. Hence answer ( C) is correct |
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| 874. |
`CO_(2)` is isostructural withA. `HgCl_2`B. `SnCl_2`C. `C_2H_2`D. `NO_2` |
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Answer» Correct Answer - A,C In `CO_2, HgCl_2, and C_2H_2` the central atoms or atoms in each molecule is `sp^2` hybridisation, as such these molecules are isostructural i.e., all are linear |
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| 875. |
The pair of species with similar shape isA. `PCl_3, NH_3`B. `CF_4, SF_4`C. `PbCl_2, CO_2`D. `PF_5, IF_5` |
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Answer» Correct Answer - A Both the molecules have pyramidal shape. |
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| 876. |
Which of the following species has a linear shape?A. `NO_2^+`B. `O_3`C. `NO_2^-`D. `SO_2` |
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Answer» Correct Answer - A `O=N^(+)=O` is linear. All other are bent molecules/ions |
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| 877. |
Which of the following species has a linear shape?A. `NO_2^+`B. `O_3` C. `NO_2^-`D. `SO_2` |
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Answer» Correct Answer - A `O=N^(+)=O` is linear. All other are bent molecules/ions |
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| 878. |
Out of `CHCl_3, CH_4 and SF_4` the molecules having regular geometry areA. `CHCl_3` onlyB. `CHCl_3 and SF_4`C. `CH_4` onlyD. `CH_4 and SF_4` |
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Answer» Correct Answer - C Regular geometry of `CH_4` is attributed to zero dipole moment `(mu = 0)` and absence of lone pair on carbon atom. |
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| 879. |
Shape of `ClF_3` isA. Equilateral triangleB. PyramidalC. V-shapedD. T-shaped |
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Answer» Correct Answer - D `ClF_3` is T-shaped |
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| 880. |
Of the three molecules `XeF_4, SF_4, SiF_4` one which have tetrahedral structures isA. all the threeB. only `SiF_4`C. both `SF_4 and XeF_4`D. only `SF_4 and XeF_4` |
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Answer» Correct Answer - C `SF_4` has seesaw shape (5 pairs of electrons) and `XeF_4` has square planar shape (6 pairs of electrons). |
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| 881. |
Molecular shape of ` SF_4, CF_4` and ` XeF_4` are :A. the same with ` 2, 0` and `1` lone pair of electron respectivelyB. the same with ` 1, 1` and `1` lone pair of electron respectivelyC. differebnt with ` 0, 1` and `2` lone pairs of electron respectivelyD. different with ` 1, 0` and `2` lone pairs of electron respectively |
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Answer» Correct Answer - D `SF_4` has ` sp^2` d-hybridisation with one lone paire , `CF_4` has `sp^3` -hybridisation with no lone pair and ` XeF_4` has `sp^4d^2` -hybridisation with two lone pairs . |
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| 882. |
Among the molecules. `SO_2`, `SF_4`, `ClF_3`. `BrF_5 and XeF_4` which of the following shape does not describe any of these moleculesA. BentB. Trigonal moleculesC. See sawD. T shape |
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Answer» Correct Answer - B `SO_2to` Bent, `SF_4 to ` Seesaw, `ClF_3 to` T shape, `BrF_5 to` square pyramidal, `XeF_4to` square planar |
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| 883. |
Which of the following statements is not true ?A. Intermolecular hydrogen bonds are formed between two different molecules of compounds.B. Intramolecular hydrogen bonds are formed between two different molecules of the same compounds.C. Intramolecular hydrogen bonds are formed within the same molecule.D. Hydrogen bonds have strong influence on the physical properties of a compound. |
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Answer» Correct Answer - B Intramolecular hydrogen bonds are formed within the same molecule. |
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| 884. |
`NaCl_((aq))` gives a white precipitate with `AgNO_(3)` solution but `C Cl_(4) or CHCl_(3)` does not , becauseA. NaCl is a covalent compound and forms AgCl as white ppt.B. NaCl is an ionic compound and forms AgCl as white ppt.C. `C Cl_(4) and CHCl_(3)` are ionic compound.D. none of these. |
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Answer» Correct Answer - B NaCl being an ionic compound and thus reacts with `AgNO_(3(aq))` to give ionic reaction , forming AgCl as white ppt. `C Cl_(4) and CHCl_(3)` being covalent do not furnish `Cl^(-)` ions in solution. `NaCl + AgNO_(3) to AgCl + NaNO_(3)` |
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| 885. |
Which one of them has the highest melting point ?A. NaClB. NaFC. NaBrD. Nal |
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Answer» Correct Answer - B Size of F atom is minimum in comparison to other halogens. When combined with Na, it has largest lattice energy |
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| 886. |
Which of the following has the highest melting point ?A. `BeCl_(2)`B. `MgCl_(2)`C. `CaCl_(2)`D. `BaCl_2` |
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Answer» Correct Answer - D The value of lattice energy depends on the charges present on the two two ions and distance between them. It shall be high if charges are high and ionic radii are small. |
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| 887. |
Lattice energy of an ionic compound depedns upon :A. charge on the ion onlyB. size of the ion onlyC. packing of ions onlyD. charge on the ion and size of the ion |
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Answer» Correct Answer - D Due to greater electronegativity difference. |
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| 888. |
Which of the following compounds contains ionic bonds?A. KlB. `CH_(4)`C. diamondsD. `H_(2)` |
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Answer» Correct Answer - A Due to greater electronegativity difference. |
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| 889. |
Which of the following pair have same hybridisation :-A. `ClO_(4)^(-) & ClO_(2)^(-)`B. `SF_(4) & C Cl_(4)`C. `BF_(3) & NF_(3)`D. `CO_(2) & SO_(2)` |
| Answer» Correct Answer - A | |
| 890. |
Which shows a changes in the type of hybridisation when :A. `NH_(3)` combines with `H^(+)`B. `AlH_(3)` combines with `H^(-)`C. in both casesD. in none cases |
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Answer» Correct Answer - B `NH_3` and `NH_4^+` are both `sp^3` -hybridised , `AlH_3` is `sp^2` and `AlH_4^(-)` is `sp^3` -hybridisesd. |
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| 891. |
The electronic structure of four elements `A,B, C, D` are (a) `1s^(2)` (b) `1s^(2), 2s^(2), 2p^(2)` ( c) `1s^(2), 2s^(2), 2p^(5)` (d) `1s^(2), 2s^(2) 2p^(6)` The tendency to form electrovalent bond is largest inA. AB. BC. CD. D |
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Answer» Correct Answer - C Element C has electronic structure `1s^(2),2s^(2)2p^(5)`, it requires only one electron to complete its octet and it will form anion so it will form electrovalent bond. |
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| 892. |
The molecule `AB_n` is planar with six pairs of electrons around A in the valence shell. The value of n isA. `6`B. `3`C. `4`D. `2` |
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Answer» Correct Answer - C The presence of 6 valence shell electron pairs implies octahedral geometry. Planar shape implies two of the octahedral positions are occupied by lone pairs while others are captured by atoms. Thus, it is `AB_4E_2` molecule. |
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| 893. |
The order of stability of metal oxides isA. `Cr_(2)O_(3) lt MgO lt Al_(2)O_(3) lt Fe_(2)O_(3)`B. `Fe_(2)O_(3) lt Cr_(2)O_(3) lt Al_(2)O_(3) lt MgO`C. `Fe_(2)O_(3) lt Al_(2)O_(3) lt Cr_(2)O_(3) lt MgO`D. `Al_(2)O_(3) lt MgO lt Fe_(2)O_(3) lt Cr_(2)O_(3)` |
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Answer» Correct Answer - B As stability is directly related to lattice energy and lattice energy depends on charge and size of ion. So, the order is `Fe_2O_(3) lt Cr_(2)O_(3) lt Al_2O_(3) lt MgO` |
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| 894. |
In which of the following moleucles are all the bonds not equal ?A. `AlF_(3)`B. `NF_(3)`C. `ClF_(3)`D. `BF_(3)` |
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Answer» Correct Answer - C `Cl` in `ClF_3` has `sp^3` d-hybridisation and possesses tow axial ` Cl-F` bonds and one at equatorial position . Two lone paires on axial position give rise to bent T shape to `ClF_3`. |
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| 895. |
The correct order of increasing electropositive character among `Cu` . Fe and ` Mg` is :A. `Cu~~ Fe gt Mg`B. `Fe gt Cu gt Mg`C. `Fe gt Mg gt Cu`D. `Mg gt Fe gt Cu` |
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Answer» Correct Answer - D `E_(OP)^@` order is ` Mggt Fe gt Cu`, Momre is ` E_(OP)^@` more is elctropositive character . |
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| 896. |
Which of the following oxides of nitrogen is ionic?A. `N_2O_5`B. `N_2O_3`C. `N_2O_4`D. `NO` |
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Answer» Correct Answer - A X-ray diffraction shows that solid `N_2O_5` is ionic `NO_2^(+)NO_3^(-)` (nitronium nitrate). |
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| 897. |
The shape of `NH_2^(-)` is like that ofA. `BeCl_2`B. `SnCl_2`C. `NO_2^(+)`D. `CS_2` |
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Answer» Correct Answer - B Steric number (SN) of `NH_2^(-)` is `2+2=4`. Therefore, it is angular. SN of `BeCl_2` is `2+0=2`. Therefore, it is linear. SN of `SnCl_2` is `2+1=3`. Therefore, it is angular. SN of `NO_2^(+)` is `2+0=2`. Therefore, it is linear. SN of `CS_2` is `2+0=2`. Therefore, it is linear. |
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| 898. |
` CuSO_4`. 5H_2 O` is represented as :A. ` [ Cu(H_2O)_5SO_4`B. `[Cu(H_2O)_3SO_4].2H_2O`C. `[Cu(H_2O)_4]SO_4.H_2O`D. `[Cu(H_2O)_5]SO_4` |
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Answer» Correct Answer - C In `CuSO_4`, only four `H_2O` act as ligand or co-ordination number for `Cu^(2+)` is four . |
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| 899. |
Among the following molecules, p-p overlap takes place inA. `H_(2)`B. `BeCl_(2)`C. `F_(2)`D. `HF` |
| Answer» Electronic configuration of fluorine is `1s^(2)2s^(2)2p^(5)` and has a half-filled p-orbital for overlapping. | |
| 900. |
Two elements `X and Y` have following electronic configurations. `X: 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(2)` `Y: 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(5)` The expected compound formed by combination of X and Y will be expresed asA. `XY_(2)`B. `X_(5)Y_(2)`C. `X_(2)Y_(5)`D. `XY_(5)` |
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Answer» Correct Answer - A From electronic configuration valencies of X and Y are +2 and -1 respectively so formula of compound is `XY_(2)`. |
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